** Homework Exercises, Ohm's Law,
Example One**

**I _{G} = 0 means points C and D are at the same potential and that I_{1} = I_{2}, I_{3} = I_{4}.**

**By Ohm’s law V _{AC} = I_{1}R_{1}, V_{AD} = I_{3}R_{3}, V_{CB} = I_{2}R_{2}, V_{DB} = I_{4}R_{4}.**

**As V _{AC} = V_{AD} and V_{CB} = V_{DB}, then I_{1}R_{1}= I_{3}R_{3},**

I_{2}R_{2}= I_{4}R_{4}.

**As I _{1} = I_{2} and I_{3} = I_{4}, then doing I_{1}R_{1}/ I_{2}R_{2}=I_{3}R_{3}/I_{4}R_{4} we get R_{1}= R_{2}(R_{3}/R_{4}).**

** Homework Exercises, Ohm's Law, Example Two**

**a) Total resistance between a and b.**

**First, calculate the 12, 6, and 4 ohm parallel combination R _{//1}1/R_{//1} = 1/12 + 1/6 + 1/4 = 1/2. Then R_{//1} = 2 ohm.**

**The upper branch is 3 ohms in series with R//1, 3 ohm + 2 ohm = 5 ohm.**

**This 5 ohm resistor is in parallel with the 20 ohm resistor, equivalent to a single resistor R _{//2} = 5 ohm//20 ohm: 1/R_{//2} = 1/5 + 1/20 = ¼. Then R_{//2} = 4. The total resistance between a and b is 4 ohm + 5 ohm = 9 ohms. The voltage drop or potential difference between points a and b is 60Ax9ohm = 540 V.**

**b) To obtain the individual currents and voltages in each resistor we can proceed as follows (there is no a unique method):**

**The voltage across R _{//2} is the difference between the total 540 voltage and the voltage drop in the 5 ohm resistor. This drop is 60Ax5ohm = 300 V. So the potential difference across R_{//2} is 540 V – 300 V = 240 V. **

Now we can calculate the currents and voltages in the R_{//2} combination:

In the 20 ohm resistor: current= 240 V/20 ohm = 12 A, voltage = 240 V

The upper branch has then a current of 60 A - 12 A = 48 A, or alternatively 240 V/5 ohm = 48A.

The 3 ohm resistor has a voltage of 48 Ax3 ohm = 144 V.

The R_{//1} combination presents a voltage of 48 AxR_{//1} = 48 Ax2 ohm = 96 V, or alternatively 240 V - 144 V = 96 V.

The currents in the 12 ohm, 6 ohm and 4 ohm are respectively 96/12 A=8 A, 96/6=16 A and 96/4=24 A.

** Homework Exercises, Ohm's Law, Example Three**

**R=R _{1}+R_{1}//R_{2}=R_{1}+R_{1}R_{2}/(R_{1}+R_{2}). **

The condition is R=R_{2}, so we do

R_{2}=R_{1}+R_{1}R_{2}/(R_{1}+R_{2}).

R_{1}R_{2}+R_{2}^{2}=R_{1}^{2}+R_{1}R_{2}+R_{1}R_{2}.

0= R_{2}^{2} - R_{1}R_{2} - R_{1}^{2}.

**This is a second degree equation in R _{2}, whose solution is R_{2}= R_{1}(1/2 ± )**

**Disregarding the minus sign we get R _{2} = 1.618R_{1}.**

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